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If \[x ^{2}-xy+y ^{2} = a\] and \[x ^{3} + y ^{3} = b\] then xy = ?
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use a^3+b^3 formula to find out the value of x+y first.
\[x ^{3} + y^{3} = (x + y)(x^{2}-xy+y^{2})\] b = (x +y) (a) b/a = x + y
\[xy=\frac{b^2-a^4}{3a^2}\]
sorry
\[xy=\frac{b^2-a^3}{3a}\]
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square x+y put the values of x^2+y^2 from your first equation.
sorry it is wrong again..
\[xy=\frac{b^2-a^3}{3a^2}\]
take the sq of last eq. you found..
so its x^2 + y^2 = b^2/a^2 - 2x .. im trying lol
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\[\frac{b^2}{a^2}=x^2+2xy+y^2\]
wait ill try to see my answer then ill compare it to cinar
\[x^2+y^2=a+xy\]
\[\frac{b^2}{a^2}=a+3xy\] \[\frac{b^2}{a^2}-a=3xy\]
thanks
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yw
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